1024. Magic IslandConstraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
There are N cities and N-1 roads in Magic-Island. You can go from one city to any other. One road only connects two cities. One day, The king of magic-island want to visit the island from the capital. No road is visited twice. Do you know the longest distance the king can go.
Input
There are several test cases in the inputA test case starts with two numbersNandK. (1<=N<=10000, 1<=K<=N). The cities is denoted from 1 to N. K is the capital.
The next N-1 lines each contain three numbersX,Y,D, meaning that there is a road between city-X and city-Y and the distance of the road is D. D is a positive integer which is not bigger than 1000.Input will be ended by the end of file.
Output
One number per line for each test case, the longest distance the king can go.
Sample Input3 11 2 101 3 20Sample Output20Problem Source
ZSUACM Team Member
这里要注意到，有n个点，但是只有n-1条路径，说明是树，用dfs求树的最长路径：
（用c++输入0.35s，，用c输入0.18s）
#include <iostream>#include <vector>#include <string.h>#include <cstring>#include <stdio.h>using namespace std;struct Road {int to, distance;//目的地和距离Road(int new_to, int new_distance) {//这样写为了读入方便to = new_to;distance = new_distance;}};vector<Road> roads[10001];//注意这里是二维动态数组bool vis[10001];int longest_road, n;void dfs(int from, int dis) {if (dis > longest_road)//更新最长的路径longest_road = dis;vis[from] = true;//这个点已经访问过，不重复访问for (int i = 0; i < (int)roads[from].size(); i++) {//遍历from所能连通的点并判断是否dfsif (!vis[roads[from][i].to]) {dfs(roads[from][i].to, dis + roads[from][i].distance);}}}int main() {int k, i, temp_from, temp_to, new_distance;while (cin >> n >> k) {longest_road = 0;memset(vis, false, sizeof(vis));memset(roads, 0, sizeof(roads));for (i = 0; i < n – 1; i++) {scanf("%d%d%d", &temp_from, &temp_to, &new_distance);roads[temp_from].push_back(Road(temp_to, new_distance));//注意这里两点是互通的roads[temp_to].push_back(Road(temp_from, new_distance));}dfs(k, 0);printf("%d\n", longest_road);}return 0;}
积极的人在每一次忧患中都看到一个机会，
